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\title{微分方程数值解\ 第12次作业}
\author{李之琪 22235056}

\begin{document}
\maketitle
\section*{HW1}
对leap-frog策略，有
\begin{eqnarray}
  \begin{aligned}
    v^{n+1} = v^{n-1} + 2kAD_0v^n.
  \end{aligned}
\end{eqnarray}
我们将上式写为
\begin{eqnarray}
  \begin{aligned}
    \begin{bmatrix}
 v^{n+1}\\
 v^n
\end{bmatrix} =
\begin{bmatrix}
2kAD_0& I \\
I & 0
\end{bmatrix}
\begin{bmatrix}
v^n \\
v^{n-1}
\end{bmatrix}=:
Q\begin{bmatrix}
v^n \\
v^{n-1}
\end{bmatrix}.
  \end{aligned}
\end{eqnarray}
则
\begin{eqnarray}
  \begin{aligned}
    \hat{Q}=
    \begin{bmatrix}
 2i\lambda A\sin\xi& I\\
 I & O
\end{bmatrix},
\end{aligned}
\end{eqnarray}
这里$\lambda = \dfrac{k}{h}, \xi = \omega h$。(4.2.48)给出，$\hat{Q}$的特征值$z$满足
\begin{eqnarray}
  \begin{aligned}
    \label{eq:eq1}
    \text{Det}(z^2I - 2i\lambda A\sin\xi z - I) = 0.
  \end{aligned}
\end{eqnarray}
对实矩阵$A$，存在可逆矩阵$S$，使得$S^{-1}AS = J$，$J$是Jordan阵。则$J$
的对角元素$\mu_i$是$A$的特征值，$i = 1,2,\cdots,N$。\eqref{eq:eq1}推出
\begin{eqnarray}
  \begin{aligned}
    \text{Det}&(SS^{-1}z^2ISS^{-1} - 2i\lambda SS^{-1}
    ASS^{-1}\sin\xi z - SS^{-1}SS^{-1}) = 0\\
    \Rightarrow \text{Det}&(S)\text{Det}(z^2I - 2i\lambda 
    J\sin\xi z - I)\text{Det}(S^{-1}) = 0\\
    \Rightarrow \text{Det}&(z^2I - 2i\lambda 
    J\sin\xi z - I) = 0.
  \end{aligned}
\end{eqnarray}
注意到上式等价于$z^2 - 2i\lambda \mu_i\sin\xi z - 1 = 0, i =
1,2,\cdots,N$，这$N$个方程的$2N$个解即为所求的$\hat{Q}$的特征值。至此，
我们给出以下算法：

(i)对实矩阵$A$，计算$A$的特征值$\mu_i$，$i = 1,2,\cdots,N$；

(ii)对每个$\mu_i$，求解二次方程$z^2 - 2i\lambda \mu_i\sin\xi z - 1 =
0$，得到$z^{(i)}_{1,2} = i\lambda \mu_i\sin\xi \pm
\sqrt{1-\lambda^2\mu_i^2\sin^2\xi}$，$i = 1,2,\cdots,N$；

(iii)$\{z^{(i)}_{1,2} \}_{1\le i \le N}$即为$\hat{Q}$的特征值。
\section*{4.2.5}
对Lax-Wendroff策略，有
\begin{eqnarray}
  \begin{aligned}
    v^{n+1} = v^{n} + kAD_0v^n + \frac{k^2A^2}{2}D_+D_-v^n.
  \end{aligned}
\end{eqnarray}
从而
\begin{eqnarray}
  \begin{aligned}
    \hat{Q} &= I + i\lambda A\sin\xi -
    2\lambda^2A^2\sin^2\frac{\xi}{2}\\
    &=
    \begin{bmatrix}
1-2\lambda^2\sin^2\frac{\xi}{2}& i\lambda \sin\xi & -2\lambda^2\sin^2\frac{\xi}{2} \\
i\lambda \sin\xi & 1-4\lambda^2\sin^2\frac{\xi}{2} & i\lambda \sin\xi\\
-2\lambda^2\sin^2\frac{\xi}{2}& i\lambda \sin\xi & 1-2\lambda^2\sin^2\frac{\xi}{2}
\end{bmatrix}.
  \end{aligned}
\end{eqnarray}
这里$\lambda = \dfrac{k}{h}, \xi = \omega h$。计算可知$\hat{Q}$的特征
值为
\begin{eqnarray}
  \begin{aligned}
    z_1 &= 1,\\
    z_2 &= 1 - \sqrt{2}i\lambda\sin\xi  -
    4\lambda^2\sin\frac{\xi^2}{2},\\
    z_3 &= 1 + \sqrt{2}i\lambda\sin\xi  -
    4\lambda^2\sin\frac{\xi^2}{2}.
  \end{aligned}
\end{eqnarray}
由于$\|z_1\| = 1$，故问题不是dissipative的。
\section*{4.2.6}
考虑
\begin{eqnarray}
  \begin{aligned}
    (I-kQ_1(x_j,t_{n+1}))v_j^{n+1} = 2kQ_0(x_j,t_n)v_j^n + (I+kQ_1(x_j,t_{n-1}))v_j^{n-1}.
  \end{aligned}
\end{eqnarray}
我们略去$x_j$和下标$_j$，则有
\begin{eqnarray}
  \begin{aligned}
    v^{n+1} - v^{n-1} = 2kQ_0(t_n)v^n + k(Q_1(t_{n+1})v^{n+1} + Q_1(t_{n-1})v^{n-1}).
  \end{aligned}
\end{eqnarray}
两边同乘$v^{n+1} + v^{n-1}$并取实部，得
\begin{eqnarray}
  \begin{aligned}
    &\left\|v^{n+1}\right\|^2 - \left\|v^{n-1}\right\|^2 \\
    &= 2k Re
    (v^{n+1} + v^{n-1}, Q_0(t_n)v^n) + k Re(v^{n+1} + v^{n-1}, Q_1(t_{n+1})v^{n+1} + Q_1(t_{n-1})v^{n-1}).
  \end{aligned}
\end{eqnarray}
我们只需要证明存在$\tilde{\alpha}$，使得
\begin{eqnarray}
  \begin{aligned}
    \label{eq:key}
Re(v^{n+1} + v^{n-1}, Q_1(t_{n+1})v^{n+1} + Q_1(t_{n-1})v^{n-1}) \le
\tilde{\alpha}(\left\|v^{n+1}\right\|^2 + \left\|v^{n-1}\right\|^2).
\end{aligned}
\end{eqnarray}
若\eqref{eq:key}成立，结合$k\left\|Q_0(t)\right\|_h \le 1 - \delta,
\delta > 0$对$t$一致成立，那么用与Theorem 4.2.10中一样的证明可以说明稳
定性。下面证明\eqref{eq:key}，记$A = Re(v^{n+1} + v^{n-1},
Q_1(t_{n+1})v^{n+1} + Q_1(t_{n-1})v^{n-1})$。

我们假设$A > 2Re(v^{n+1} + v^{n-1}, Q_1(t_{n+1})(v^{n+1} + v^{n-1}))$
且$A > 2Re(v^{n+1} + v^{n-1}, Q_1(t_{n-1})(v^{n+1} + v^{n-1}))$。否则，
根据$Q_1$的semi-bounded性质，\eqref{eq:key}显然成立。则有
\begin{eqnarray}
  \begin{aligned}
    \label{eq:mid1}
A &> Re(v^{n+1} + v^{n-1}, Q_1(t_{n+1})(v^{n+1} + v^{n-1})) +
Re(v^{n+1} + v^{n-1}, Q_1(t_{n-1})(v^{n+1} + v^{n-1}))\\
&\Rightarrow Re(v^{n+1} + v^{n-1}, Q_1(t_{n+1})v^{n-1} +
Q_1(t_{n-1})v^{n+1}) < 0.
\end{aligned}
\end{eqnarray}
从而
\begin{eqnarray}
  \begin{aligned}
    \label{eq:mid2}
A &= Re(v^{n+1} + v^{n-1},
Q_1(t_{n+1})v^{n+1} + Q_1(t_{n-1})v^{n-1}) \\
&= Re(v^{n+1}, Q_1(t_{n+1})v^{n+1}) + Re(v^{n-1}, Q_1(t_{n-1})v^{n-1})\\
&\quad + Re(v^{n+1}, Q_1(t_{n-1})v^{n-1}) + Re(v^{n-1},
Q_1(t_{n+1})v^{n+1})\\
&= Re(v^{n+1}, Q_1(t_{n+1})v^{n+1}) + Re(v^{n-1},
Q_1(t_{n-1})v^{n-1})\\
&\quad + Re(Q_1(t_{n-1})v^{n+1}, v^{n-1}) + Re(Q_1(t_{n+1})v^{n-1},
v^{n+1})\\
&< Re(v^{n+1}, Q_1(t_{n+1})v^{n+1}) + Re(v^{n-1},
Q_1(t_{n-1})v^{n-1})\\
&\quad + Re(v^{n+1}, -Q_1(t_{n-1})v^{n+1}) + Re(v^{n-1},
-Q_1(t_{n+1})v^{n-1})\\
&= Re(v^{n+1}, (Q_1(t_{n+1})-Q_1(t_{n-1}))v^{n+1}) + Re(v^{n-1},
(Q_1(t_{n-1})-Q_1(t_{n+1}))v^{n-1})\\
&< 2k\alpha (\left\|v^{n+1}\right\|^2 + \left\|v^{n-1}\right\|^2) =: \tilde{\alpha}(\left\|v^{n+1}\right\|^2 + \left\|v^{n-1}\right\|^2).
\end{aligned}
\end{eqnarray}
这里第三步实际上假设了$Q_1$是Hermite阵，第四步用到\eqref{eq:mid1}，第
六步假设了$Q_1$关于$t$的一致Lipschitz连续性。基于实际应用和问题的适定
性，我们认为以上假设是合理的。
\end{document}

